Question

Consider the following balanced equation:

3CuO(s) + 2NH3(g) → 3H2O(l) + 3Cu(s) + N2(g)
If 70.8 moles of CuO(s) and 40.1 moles of NH3(g) are allowed to react, what is the theoretical yield of Cu(s) in moles?

Answer 1

The theoretical yield of Cu(s) in moles is 60.15 moles

Step-by-step explanation:

Step 1: Data given

Number of moles CuO = 70.8 moles

Number of moles NH3 = 40.1 moles

Molar mass CuO = 79.545 g/mol

Molar mass NH3 = 17.03 g/mol

Step 2: The balanced equation

3CuO(s) + 2NH3(g) → 3H2O(l) + 3Cu(s) + N2(g)

For 3 moles CuO we need 2 moles NH3 to produce 3 moles H2O, 3 moles Cu and 1 mol N2

NH3 is the limiting reactant. It will completely be consumed (40.1 moles). CuO is in excess. There will react 3/2 * 40.1 = 60.15 moles

There will remain 70.8 - 60.15 = 10.65 moles CuO

Step 3: Calculate moles Cu

For 3 moles CuO we need 2 moles NH3 to produce 3 moles H2O, 3 moles Cu and 1 mol N2

For 40.1 moles NH3 we'll have 60.15 moles Cu

The theoretical yield of Cu(s) in moles is 60.15 moles